SOLUTION: A 20-ft diagonal brace on a bridge connects a support of the center of the bridge to a side support on the bridge. The horizontal distance that it spans is 4 ft longer than the hei

Question 548519: A 20-ft diagonal brace on a bridge connects a support of the center of the bridge to a side support on the bridge. The horizontal distance that it spans is 4 ft longer than the height that it reaches on the side of the bridge. Find the horizontal and vertical distances spanned by this brace.
Found 2 solutions by stanbon, oberobic:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
A 20-ft diagonal brace on a bridge connects a support of the center of the bridge to a side support on the bridge. The horizontal distance that it spans is 4 ft longer than the height that it reaches on the side of the bridge. Find the horizontal and vertical distances spanned by this brace.
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Draw the picture:
You have a right triangle with:
hypotenuse = 20 ft
height = x
base = x+4
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Equation:
x^2 + (x+4)^2 = 20^2
x^2 + x^2 + 8x + 8 = 400
----
2x^2 + 8x - 392 = 0
x^2 + 4x - 196 = 0
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I graphed the quadratic and found x = 12.14 ft (height)
base = x+4 = 16.14 ft.
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You could use the quadratic formula.
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Cheers,
Stan H.
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Answer by oberobic(2304) (Show Source): You can put this solution on YOUR website!
x = height
x+4 = width
20 = hypotenuse
.
20^2 = x^2 + (x+4)^2
.
400 = x^2 + x^2 + 8x + 16
.
0 = 2x^2 +8x - 384
.
2x^2 +8x - 384 = 0
.
x^2 + 4x -192 = 0
.
factor
.
(x-12)(x+16) = 0
.
zeroes of x = 12 or -16
.

.
x = -16 is not applicable
.
x = 12 is the height of the span
x+4 = 16 is the width of the span
.
Check using the Pythagorean formula.
12^2 + 16^2 = 144+256 = 400
20^2 = 400
Correct.
.
Done.
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